A) tan A tan B < 1
B) tan A tan B > 1
C) tan A tan B = 1
D) None of these
Correct Answer: A
Solution :
Since, angle C is obtuse, therefore angles A and B are acute \[\therefore \] \[\tan C>0,\,\tan A<0,\,\tan \,B<0\] Now, \[\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\] \[\Rightarrow \]\[\tan (\pi -C)=\frac{\tan A+\tan B}{1-\tan A\tan B}\] \[\Rightarrow \]\[-\tan C=\frac{\tan \,A+\tan B}{1-\tan A\tan B}\] \[\Rightarrow \]\[1-\tan A\tan B>0\] (\[\because \] LHS and numerator are \[+ve\] \[\Rightarrow \]\[\tan \,A\tan B<1\]
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